leetcode 25.Reverse Nodes in k Group

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  Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

  Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

Only constant extra memory is allowed.
You may not alter the values in the list’s nodes, only nodes itself may be changed.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-nodes-in-k-group
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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* cur = head;
        return reverse(head, cur, k);
    }

    ListNode* reverse(ListNode* head, ListNode* cur, int k){
        ListNode* tmp = cur;
        int count = 0;
        while(tmp){
            tmp  = tmp -> next;
            count++;
        }
        if(count < k) return head;

        ListNode* next = NULL;
        ListNode* pre = NULL;
        int n = k;

        while(cur && n-- > 0){
            next = cur -> next;
            cur -> next = pre;
            pre = cur;
            cur = next;
        }

        head -> next = reverse(cur, cur, k);

        return pre;
    }
};